Question

$\sqrt{3}\text{cosec}{20}^{\circ }-\sec {20}^{\circ }$=

• A. 2
• B. $\frac{2sin{20}^{\circ }}{sin{40}^{\circ }}$
• C. 4
• D. $\frac{4sin{20}^{\circ }}{sin{40}^{\circ }}$

Answer 1

The correct option is C

4

$\sqrt{3}\text{cosec}{20}^{\circ }-\sec {20}^{\circ }=\frac{\sqrt{3}}{\sin {20}^{\circ }}-\frac{1}{\cos {20}^{\circ }}$
$=\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }}=\frac{2[\frac{\sqrt{3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ }]}{\frac{2}{2}\sin {20}^{\circ }\cos {20}^{\circ }}$
$=\frac{4\cos ({20}^{\circ }+{30}^{\circ })}{\sin {40}^{\circ }}=\frac{4\cos {50}^{\circ }}{\sin {40}^{\circ }}=\frac{4\sin {40}^{\circ }}{\sin {40}^{\circ }}=4$.